If we delete two points $x,y$ from $\mathbb{A}^1$, can we without loss of generality assume $x=0, y=1$?

Question

My intuition is that we can assume this. More precisely, what I mean is, suppose $\mathbb{A}^1_k$ is the affine space over an algebraically closed field $k$. If $x,y$ are any two distinct points in $k$, is it true that $\mathbb{A}^1_k\setminus\{0,1\}\cong\mathbb{A}^1_k\setminus\{x,y\}$ are isomorphic?

Answer 1

Yes, it is true. Define $\varphi:\mathbb{A}^1\to\mathbb{A}^1$ by $$p\mapsto\frac{p-x}{y-x},$$and note that $\varphi(x)=0,\;\varphi(y)=1$.

Answer 2

Just find an affine transform which maps two points to any given other two points. If you dont want to work (joint the club) remember that the collineations of the projective plane are (sharply) three transitive and of you fix the point at infinity to get the affine plane we see that the affine transforms are sharply 2-transitive.